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IRISSAK [1]
3 years ago
6

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. The prob

ability of a player weighing more than 250 pounds is Group of answer choices 0.9505 0.9772 0.0228 0.4505 0.0495
Mathematics
1 answer:
stepan [7]3 years ago
4 0

Answer:

0.0228

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

This means that \mu = 200, \sigma = 25

The probability of a player weighing more than 250 pounds is

This is 1 subtracted by the pvalue of Z when X = 250. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{250 - 200}{25}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

The probability of a player weighing more than 250 pounds is 0.0228

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3 years ago
Scores from an exam are normally distributed with a mean of 85 and a standard deviation of 5. Suppose 188 students take the exam
charle [14.2K]

The mean is \mu=85 and standard deviation is \sigma=5. Then the variable X\sim N(85,5).

Use substitution Z=\dfrac{X-\mu}{\sigma}=\dfrac{X-85}{5}, then the variable Z\sim N(0,1).

The probability Pr(X>80) can be calculated in the following way:

for X=80, Z=\dfrac{80-85}{5} =-1 and Pr(X>80)=Pr(Z>-1). From the diagram Pr(Z>-1)=0.34+0.34+0.135+0.025=0.84.

Conclusion: if 188 students take the exam, then 188·0.84=157.92 (157) would receive a score above 80

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3 years ago
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3 years ago
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How to do the problem
marta [7]
There are 153 people who each ate 1/4 pounds. Therefore, the amount of watermelon eaten is 153/4 pounds.
The amount of watermelon served is
8 \frac{1}{4}  + 9 \frac{1}{4}  + 8 \frac{7}{8}  + 9 \frac{5}{8}  + 10 \frac{3}{4}
=  \frac{8 \times 4 + 1}{4}  +  \frac{9 \times 4 + 1}{4}  \\  +  \frac{8 \times 8 + 7}{8}  +  \frac{9 \times 8 + 5}{8} \\  +  \frac{10 \times 4 + 3}{4}
=  \frac{33}{4}  +  \frac{37}{4}  +  \frac{71}{8}  +  \frac{77}{8}  +  \frac{43}{4}  \\  =  \frac{33 \times 2}{4 \times 2}  +  \frac{37 \times 2}{4 \times 2}  +  \frac{71}{8}  \\  +  \frac{77}{8}  +  \frac{43 \times 2}{4 \times 2}
=  \frac{66 + 74 + 71 + 77 + 86}{8}
=  \frac{374}{8}  =  \frac{186}{4}
There is 186/4 total watermelon. Minus the 153/4 that was eaten,
\frac{186 - 153}{4}  =  \frac{33}{4}  = 8 \frac{1}{4}

The leftover watermelon is 8 & 1/4 pounds.
6 0
3 years ago
On a number line, what is the distance between -29 and 100? A) 29 B) 71 C) 81 D) 129
shusha [124]

-29 is 29 bars away from 0. 100 is 100 bars away from 0.

29 + 100 = 129

So the distance between -29 and 100 is 129. The answer is D.

6 0
3 years ago
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